If k_f=1.86*""^(@)C*mol^-1*kg for water, what is the freezing point of a solution composed of 255*kg CaCl_2, and 33.8*L of water?

1 Answer
anor277
Apr 27, 2017

Well, DeltaT_"fusion"=k_fxxcxxi where.................

Explanation:

Where k_f is the "molal freezing point depression constant", which you have kindly given, c="molal concentration" in mol*kg^-1, and i, the so-called "van't Hoff factor" which reflects the speciation of an ionic solid in a solvent.

Clearly, here, i=3. Why? Because in solution, "calcium chloride" speciates to give three ions......

CaCl_2(aq) stackrel(H_2O)rarrCa^(2+) + 2Cl^(-)

And so...............

Delta_"fusion"=1.86*""^(@)C*m^-1xx3xx((2.55xx10^3*g)/(110.98*g*mol^-1))/(33.8*kg)=3.80 ""^@C.

This reflects the depression of the fusion point (the melting point) with RESPECT to PURE SOLVENT. And thus "fusion point" -= -3.80 ""^@C.

In practice, the melting point probably would not be depressed so far. And this is the province of measurement.

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